math
posted by becky .
the vertices are at (2,1) and (2,7) and focus is at (2,*)
write the equation of the hyperbola that meets each set of conditions.
Is the asterisk supposed to be an 8? It has to be some number. I will set this up for you later of you respond. The fomulas you need are in most calculus and analytic geometry texts.
yeah it is an 8
your vertices are (2,1) and (2,7), so your centre is (2,4), the midpoint.
Since your hyperbola is vertical it must have equation:
(x2)^2 over a^2  (y4)^2 over b^2 = 1
the distance from (2,4) to (2,7) is 3, so b=3
the distance from (2,4) to your focus (2,8) is 4, so c=4
recall that in a hyperbola a^2 + b^2 = c^2, substitute to get a^2 = 7
Therefore (x2)^2 over 7  (y4)^2 over 9 = 1
Nice job, reiny! Thanks for helping us out here.
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